QPaperGen

We have:
\[
f(x) = (x + 1)^3(x - 3)^3
\]

Differentiate using product rule:
\[
f'(x) = 3(x + 1)^2(x - 3)^3 + 3(x - 3)^2(x + 1)^3
\]

Factor common terms:
\[
f'(x) = 3(x + 1)^2(x - 3)^2[(x - 3) + (x + 1)]
\]
\[
= 3(x + 1)^2(x - 3)^2(2x - 2)
\]
\[
= 6(x + 1)^2(x - 3)^2(x - 1)
\]

Now, set derivative equal to zero:
\[
f'(x) = 0 \Rightarrow (x + 1)^2 = 0 \text{ or } (x - 3)^2 = 0 \text{ or } (x - 1) = 0
\]
\[
\Rightarrow x = -1,\ 1,\ 3
\]

These values divide the number line into intervals:
\[
(-\infty, -1),\ (-1, 1),\ (1, 3),\ (3, \infty)
\]

Now we analyze the sign of \(f'(x)\) in each interval:
\[\]

In \((- \infty, -1)\), choose \(x = -2\):
\(f'(x) = 6(-1)^2(-5)^2(-3) < 0\)
So, \(f(x)\) is \(\textbf{strictly decreasing}\).

\[\]

In \((-1, 1)\), choose \(x = 0\):
\(f'(x) = 6(1)^2(-3)^2(-1) < 0\)
So, \(f(x)\) is \(\textbf{strictly decreasing}\).

\[\]

In \((1, 3)\), choose \(x = 2\):
\(f'(x) = 6(3)^2(-1)^2(1) > 0\)
So, \(f(x)\) is \(\textbf{strictly increasing}\).

\[\]

In \((3, \infty)\), choose \(x = 4\):
\(f'(x) = 6(5)^2(1)^2(3) > 0\)
So, \(f(x)\) is \(\textbf{strictly increasing}\).

\[\]

Conclusion:
\[
f(x) \text{ is strictly decreasing in } (-\infty, -1) \cup (-1, 1)
\]
\[
f(x) \text{ is strictly increasing in } (1, 3) \cup (3, \infty)
\]